Question

If $O$ is the circumcentre of the $\Delta ABC$ and $R_1,R_2$ and $R_3$ are the radii of the circumcircles $OBC,OCA$ and $OAB$ repectively then $\frac{a}{R_1}+\frac{b}{R_2}+\frac{c}{R_3}$ has the value equal to:

• A. $\frac{abc}{2R^3}$
• B. $\frac{abc}{R^3}$
• C. $\frac{4\Delta }{R^2}$
• D. $\frac{\Delta }{4R^2}$

Answer 1

The correct option is B $\frac{abc}{R^3}$

In $△OBC$

$\angle BOC=2A$

From sine rule:$\frac{BC}{\sin \angle BOC}=2R_1$

$\Rightarrow \frac{a}{R_1}=2\sin 2A$

In $△OAC$

$\angle AOC=2B$

From sine rule:$\frac{AC}{\sin \angle AOC}=2R_2$

$\Rightarrow \frac{b}{R_2}=2\sin 2B$

In $△OAB$

$\angle AOB=2C$

From sine rule:$\frac{AB}{\sin \angle AOB}=2R_3$

$\Rightarrow \frac{c}{R_3}=2\sin 2C$

Now,$y=\frac{a}{R_1}+\frac{b}{R_2}+\frac{c}{R_3}=2(\sin 2A+\sin 2B+\sin 2C)$

$\Rightarrow y=2\left(2\sin \left(\frac{2A+2B}{2}\right)\cos \left(\frac{2A-2B}{2}\right)\right)+4\sin C\cos C$

$\Rightarrow y=4\sin (\pi -C)\cos (A-B)+4\sin C\cos (\pi -(A+B))$

$\Rightarrow y=4\sin C\cos (A-B)-4\sin C\cos (A+B)$

$\Rightarrow y=4\sin C[\cos (A-B)-\cos (A+B)]$

$\Rightarrow y=4\sin C[-2\sin \left(\frac{A-B+A+B}{2}\right)\sin \left(\frac{A-B-A-B}{2}\right)]$

$\Rightarrow y=4\sin C\left[2\sin A\sin B\right]$

$\Rightarrow y=8\sin A\sin B\sin C$

$\Rightarrow y=8\times \frac{a}{2R_1}\times \frac{b}{2R_2}\times \frac{c}{2R_3}$

$\Rightarrow y=8\times \frac{abc}{8R_1{R}_2{R}_3}$

$\Rightarrow y=8\times \frac{abc}{8R^3}$ where $R_1=R_2=R_3=R$

$\therefore y=\frac{abc}{R^3}$